This is a continuum to this question which I seem to not understand the solution.
Integral around unit sphere of inner product
I don't get how the diagonal is done, I get that (for example, for $n=3$ that) $$\iiint_B \langle Av, v \rangle dxdydz = \iiint_B a_{11} x_1^2 + a_{22} x_2^2 + a_{33} x_3^2 dx_1dx_2 dx_3 =$$ $$ = (a_{11} + a_{22} + a_{33}) \iint_B x_1^2dx_1dx_2dx_3$$ Which according to the solution is just $\frac{1}{3} Tr(A) \cdot Area(B)$
The part I fail to see is just how is $$\iiint_Bx_1^2dx_1dx_2dx_2=\frac{1}{3}Area(B)$$
By symmetry, $$\iiint_B x_1^2\,dx_1\,dx_2\,dx_3 =\iiint_B x_2^2\,dx_1\,dx_2\,dx_3 =\iiint_B x_3^2\,dx_1\,dx_2\,dx_3 $$ so that \begin{align} \iiint_B(a_{11}x_1^2+a_{22}x_2^2+a_{33}x_3^2)\,dx_1\,dx_2\,dx_3 &=(a_{11}+a_{22}+a_{33})\iiint_B x_1^2\,dx_1\,dx_2\,dx_3\\ &=\frac{a_{11}+a_{22}+a_{33}}3\iiint_B(x_1^2+x_2^2+x_3^2)\,dx_1\,dx_2\,dx_3\\ &=\frac{\text{Tr}(A)}3\iiint_B(x_1^2+x_2^2+x_3^2)\,dx_1\,dx_2\,dx_3. \end{align} Now we can evaluate $$\iiint_B(x_1^2+x_2^2+x_3^2)\,dx_1\,dx_2\,dx_3$$ by spherical polars...